Question: Let $f(x, y, z) = \dfrac{xy}{z}$ Suppose $\vec{a} = (-3, 2, 1)$ and $\vec{v} = \left( 1, 1, 1 \right)$. Find the directional derivative of $f(x, y, z)$ at $\vec{a}$ in the direction of $\vec{v}$. Do not normalize the direction vector for your calculation. $\dfrac{\partial f}{\partial v} = $
The directional derivative of a function $f$ at $\vec{a}$ in the direction of $\vec{v}$ equals $\left( \nabla f (\vec{a}) \right) \cdot \vec{v}$. Let's find the gradient of $f$. $\begin{aligned} \nabla f &= \left( \dfrac{\partial f}{\partial x}, \dfrac{\partial f}{\partial y}, \dfrac{\partial f}{\partial z} \right) \\ \\ &= \left( \dfrac{y}{z}, \dfrac{x}{z}, -\dfrac{xy}{z^2} \right) \end{aligned}$ Plugging in $(-3, 2, 1)$, we can find $\nabla f$ at $\vec{a}$. $\nabla f (\vec{a}) = (2, -3, 6)$ Therefore: $\begin{aligned} \left( \nabla f (\vec{a}) \right) \cdot \vec{v} &= (2, -3, 6) \cdot (1, 1, 1) \\ \\ &= 5 \end{aligned}$ In conclusion, the directional derivative of $f$ at $\vec{a}$ in the direction of $\vec{v}$ equals $5$.